Find $\lim_{x\to \scriptsize\dfrac{\pi}{4}}\dfrac{\cos(2x)}{\cos(x)-\sin(x)}$. Choose 1 answer: Choose 1 answer: (Choice A) A $\sqrt{2}$ (Choice B) B $2$ (Choice C) C $4$ (Choice D) D The limit doesn't exist
Explanation: Substituting $x=\dfrac{\pi}{4}$ into $\dfrac{\cos(2x)}{\cos(x)-\sin(x)}$ results in the indeterminate form $\dfrac{0}{0}$. This doesn't necessarily mean the limit doesn't exist, but it does mean we have to work a little before we find it. Since our expression includes trigonometric functions, let's try to re-write it using factorization and trigonometric identities. Since we have $\cos(2x)$ in our expression, let's rewrite it using one of its double angle identities. Since the expression in the denominator includes $\cos(x)-\sin(x)$, the most fitting identity is $\cos(2x)=\cos^2(x)-\sin^2(x)$. $\begin{aligned} &\phantom{=}\dfrac{\cos(2x)}{\cos(x)-\sin(x)} \\\\ &=\dfrac{\cos^2(x)-\sin^2(x)}{\cos(x)-\sin(x) } \gray {\text{cos(2x) identity}} \\\\ &=\dfrac{(\cos(x)-\sin(x))(\cos(x)+\sin(x))}{\cos(x)-\sin(x) } \gray{\text{Diff. of squares}} \\\\ &=\dfrac{\cancel{(\cos(x)-\sin(x))}(\cos(x)+\sin(x))}{\cancel{\cos(x)-\sin(x)} } \gray{\text{Cancel common factors}} \\\\ &=\cos(x)+\sin(x)\text{, for }x\neq \{...,-\dfrac{3}{4}\pi, \dfrac{1}{4}\pi, \dfrac{5}{4}\pi, \dfrac{9}{4}\pi,...\} \end{aligned}$ This means that the two expressions have the same value for all $x$ -values (in their domains) except for $-\dfrac{3}{4}\pi+k\pi $ for any integer $k$, and specifically $\dfrac{\pi}{4}$. We can now use the following theorem: If $f(x)=g(x)$ for all $x$ -values in a given interval except for $x=c$, then $\lim_{x\to c}f(x)=\lim_{x\to c}g(x)$. In our case, $\dfrac{\cos(2x)}{\cos(x)-\sin(x)}=\cos(x)+\sin(x)$ for all $x$ -values in the interval $(0,\dfrac{\pi}{2})$ except for $x=\dfrac{\pi}{4}$. Therefore, $\lim_{x\to \scriptsize\dfrac{\pi}{4}}\dfrac{\cos(2x)}{\cos(x)-\sin(x)}=\lim_{x\to \scriptsize\dfrac{\pi}{4}}\cos(x)+\sin(x)=\sqrt{2}$ (The last limit was found using direct substitution.) In conclusion, $\lim_{x\to \scriptsize\dfrac{\pi}{4}}\dfrac{\cos(2x)}{\cos(x)-\sin(x)}=\sqrt{2}$.